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\begin{document}

\title{高等代数二}
\subtitle{04-实系数-复系数-有理系数多项式}
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{{\ppr LQW}}
\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
%\date{{\ppr 2023年3月9日} }

\maketitle

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\begin{frame}{4.1. 作业：星期天晚上十点半之前在网络教学平台提交 }

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\begin{enumerate}
\item   整理课堂笔记里的重点难点。补充没写完的计算或证明。
\item   习题(2.7)任选2题。抄写题目。
\item   习题(2.8)任选2题。抄写题目。
\end{enumerate}

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\begin{enumerate}

\item   代数基本定理
\item   韦达定理
\item   复数域上的不可约多项式
\item   实数域上的不可约多项式
\item   求根公式

\item   有理系数多项式
\item   整系数多项式
\item   高斯引理
\item   Eisentein判别法
\item   整系数多项式的有理根

\end{enumerate}

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\begin{enumerate}

\item   在实数范围和复数范围内分解因式。
\item   多项式的根与系数的关系。
\item   计算整系数多项式的有理根。
\item   有理系数多项式与整系数多项式的关系。
\item   用Eisenstein判别法判断有理系数多项式是否不可约。

%\item   

\end{enumerate}

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\begin{itemize}

\item  {\color{red}定义：所有复数组成的集合，以及加减乘除四则运算，符合``域''的所有公理。这个代数系统称为复数域。}

\item  {\color{red}问题：简述复数的两种表示及其相互关系。}

\item  解答：用直角坐标和极坐标，复数可以写成下述方式，
\begin{align*}
x = a+bi = re^{i\theta}. 
\end{align*}
它们之间的关系是
\begin{align*}
\left\{\begin{array}{rcl}
a &=& r\cos\theta, \\ 
b &=& r\sin\theta.
\end{array}\right.
\hspace{1cm}
\left\{\begin{array}{rcl}
r &=& \sqrt{a^2+b^2}, \\ 
\tan\theta &=& \frac{y}{x}.
\end{array}\right.
\end{align*}



\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.7.1. 代数基本定理：任意次数大于等于1的多项式，在复数范围内至少有1个根。}


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：在复数范围内将 $f(x)=x^4-2$ 分解因式。}

\item  解答：记 $a=\sqrt[4]{2}$ 以及 $i=\sqrt{-1}$, 则 $$x^4-2 = (x-a)(x-ai)(x+a)(x+ai).$$

\item  注：多项式 $f(x)=x^4-2$ 的四个复数根均匀地落在半径为 $a$ 的圆周上。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：在复数范围内分解因式：}
\begin{enumerate}
\item {\color{red} $f(x)=x^3+2$. }
\item {\color{red} $f(x)=x^3+8$. }
\end{enumerate}

\item  解答：考虑 $f(x)=x^3+b$, 其中 $b$ 是一个给定的复数。设 $x=ay$, 其中 $a^3=b$. 则
\begin{align*} 
f(x) &= x^3+b \\
&= (ay)^3+b \\
&= b(y^3+1).
\end{align*}
因此这个问题可以化为在复数范围内对多项式 $g(y) = y^3+1$ 分解因式。

\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.7.2. 任意 $n\,\, (n\ge 1)$ 次多项式在复数范围内有 $n$ 个根，重根按重数计算。}

\item  证明：由代数基本定理，$f(x)$ 至少有1个复根 $\alpha_1$. 于是
$$f(x)=(x-\alpha_1)f_1(x),$$
这里 $f_1(x)$ 是一个 $n-1$ 次多项式。对 $f_1(x)$ 继续应用代数基本定理，最后可得
$$f(x)=(x-\alpha_1)(x-\alpha_2)\cdots (x-\alpha_n).$$


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\begin{itemize}

\item  {\color{red}推论：复数域上的不可约多项式正好是那些一次多项式。}

\item  问：其它数域上的不可约多项式是那些呢？

\item  答：其它数域上的不可约多项式的情况就比较复杂。

\end{itemize}

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\begin{itemize}

\item  {\color{red}韦达定理：是描述多项式的根与系数的关系的一个定理。设 
\begin{eqnarray*}
f(x) = a_0x^n+a_1x^{n-1}+\cdots+a_n 
\end{eqnarray*}
的根分别是 $x_1, x_2, \cdots, x_n$. 则有下述等式
\begin{eqnarray*}
\frac{a_1}{a_0} &=& - (x_1+x_2+\cdots+x_n), \\
\frac{a_2}{a_0} &=& x_1x_2+x_1x_3+\cdots+x_{n-1}x_n, \\
\frac{a_3}{a_0} &=& - (x_1x_2x_3+x_1x_2x_4+\cdots+x_{n-2}x_{n-1}x_n), \\
\vdots&&\vdots \\
\frac{a_n}{a_0} &=& (-1)^{n} x_1x_2\cdots x_n. 
\end{eqnarray*}
}

\end{itemize}

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\begin{itemize}

\item  韦达定理的证明：因为 $x_1,x_2,\cdots,x_n$ 都是 $f(x)$ 的根，所以有 
\begin{eqnarray*}
f(x) &=& a_0x^n+a_1x^{n-1}+\cdots+a_n \\ &=& a_0(x-x_1)(x-x_2)\cdots (x-x_n). 
\end{eqnarray*}
将这些乘积乘出来，然后比较系数可得。


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\begin{itemize}

\item  {\color{red}例子：设 $f(x)=x^3+ax^2+bx+c$ 的三个根分别为 $x_1,x_2,x_3$. 
求一个多项式，以 $kx_1,kx_2,kx_3$ 为根。}

\item  解答：由条件可得 
\begin{eqnarray*}
f(x) &=& x^3+ax^2+bx+c = (x-x_1)(x-x_2)(x-x_3) \\
&=&  x^3 - (x_1+x_2+x_3)x^2 + (x_1x_2+x_1x_3+x_2x_3)x - x_1x_2x_3.
\end{eqnarray*}
于是所求多项式为  
\begin{eqnarray*}
g(x) &=& (x-kx_1)(x-kx_2)(x-kx_3) \\ 
&=& x^3 -k(x_1+x_2+x_3)x^2 + k^2(x_1x_2+x_1x_3+x_2x_3)x - k^3x_1x_2x_3 \\ 
&=& x^3 + kax^2 + k^2bx + k^3c.
\end{eqnarray*}

\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.7.3. 设 $f(x)\in \mathbb{R}[x]$ 是一个实系数多项式。设 $\alpha$ 是 $f(x)$ 的一个非实数的复数根，则它的共轭 $\overline{\alpha}$ 也是 $f(x)$ 的根。 即实系数多项式的复根是成对出现的。}

\item  证明：由复数的共轭运算保持加减乘除四则运算可知，
\begin{eqnarray*}
a_0x^n+a_1x^{n-1}+\cdots+a_n &=& 0, \\
\Leftrightarrow \overline{a_0x^n+a_1x^{n-1}+\cdots+a_n} &=& 0, \\
\Leftrightarrow a_0\overline{x}^n+a_1\overline{x}^{n-1}+\cdots+a_n &=& 0.
\end{eqnarray*}
其中用到这个多项式的系数都是实数这个条件。


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\begin{itemize}

\item  {\color{red}问题：求下述方程的所有复数根。}
\begin{enumerate}
\item  {\color{red} $x^2+2x+2=0$. } 
\item  {\color{red} $x^4+2x^2+2=0$. } 
\end{enumerate}

\item  解答：配方法。


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\begin{itemize}

\item  {\color{red}定理2.7.4. 实数域上的不可约多项式如下：}
\begin{enumerate}
\item   {\color{red}一次多项式 $a_0x+a_1,\quad a_0\neq 0$. }
\item   {\color{red}二次多项式 $a_0x^2+a_1x+a_2,\quad a_0\neq 0, a_1^2-4a_0a_2<0$. }
\end{enumerate}

%\item  问：为什么这些是不可约的？还有其它不可约的多项式吗？

\item  证明：因为这样的二次多项式没有实数根，所以这些是不可约多项式。在实数域上没有其它形式的不可约的多项式。一种证明的思路是先在复数范围内分解因式，然后使用复根成对出现这个结论。



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\begin{itemize}

\item  {\color{red}例子：在实系数范围内分解因式 $f(x)=x^4+1$.}

\item  解答一：观察到 $f(x)=(x^2+1)^2 - 2x^2$. 然后使用平方差公式。

\item  解答二：先在复数范围内分解因式，可得
\begin{align*}
f(x) = (x-\omega)(x-\omega^3)(x-\omega^5)(x-\omega^7),
\end{align*}
其中 
\begin{align*}
\omega = \exp\left(\frac{\pi}{4}i\right) = \cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i. 
\end{align*}
然后再把共轭的因式相乘，得到实系数范围内的因式分解。


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\begin{itemize}

\item  {\color{red}定理2.7.5. 每个次数大于等于1的实系数多项式都可以分解成一些一次和二次的不可约多项式的乘积。}

\item  证明：由上一个定理得证。

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\begin{itemize}

\item  {\color{red}问题：在实数范围内分解因式。}
\begin{enumerate}
\item  {\color{red}  $f(x) = x^2+2x+2$. } 
\item  {\color{red} $f(x) = x^4+2x^2+2$. }
\end{enumerate}

\item  解答：先在复数范围内分解因式，然后合并共轭的因式。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问：关于多项式的根的研究，有哪些重要的问题？}

\item  答：

\begin{enumerate}

\item  多项式的根的近似求法，数值计算等研究。
\item  多项式的根的根式求法，近世代数等研究。
\item  多项式在不同数域范围内的根的个数，数论等研究。

\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}问：写出下述方程的求根公式，或求根方法。
\begin{eqnarray*}
f(x) &=& ax^2+bx+c \\
f(x) &=& ax^3+bx^2+cx+d \\
f(x) &=& ax^4+bx^3+cx^2+dx+e \\
f(x) &=& ax^5+bx^4+cx^3+dx^2+ex+k
\end{eqnarray*}
}

\item  答：第一个公式据说已有两千年历史， $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. $$
其它的，有点难。

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\begin{itemize}

\item  {\color{red}定义：一个整系数多项式如果它的系数是互素的，则称为本原多项式。}

\item  {\color{red}问题：判断多项式 $f(x)=5x^3+2x^2+15x+ 8$ 是不是一个本原多项式。}

\item  解答：是的。

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\begin{itemize}

\item  {\color{red}问：有理系数多项式与整系数多项式的关系如何？}

\item  解答： 挺接近的，只差有理系数因子。例如
\begin{eqnarray*}
f(x) &=& \frac{3}{2}x^3+ \frac{3}{5}x^2+ \frac{9}{2}x+ \frac{12}{5} \\
&=& \frac{3}{10}( 5x^3+2x^2+15x+ 8)
\end{eqnarray*}
即：有理系数多项式可以写成一个有理数与一个本原多项式的乘积。

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\begin{itemize}

\item  {\color{red}引理2.8.1.(高斯引理) 两个本原多项式的乘积仍然是一个本原多项式。}

\item  证明： 将两个本原多项式写出来，将乘积算出来，证明系数没有公因子。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：举例说明高斯引理。}

\item  解答：下述多项式 $f(x)$ 和 $g(x)$ 是本原多项式，乘积 $f(x)g(x)$ 也是多项式。
\begin{eqnarray*}
f(x) &=& 2x+3, \\
g(x) &=& 3x^2+4x+6, \\
f(x)g(x) &=& 6x^3+17x^2+24x+18.
\end{eqnarray*}

\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.8.1. 如果一个整系数多项式在有理数域上可约，那么它也能写成两个次数较低的整系数多项式的乘积。}

\item  证明： 
\begin{enumerate}
\item  设 $f(x)=g_1(x)g_2(x)$, 其中 $g_1(x)$ 和 $g_2(x)$ 是两个有理系数多项式。
\item  按照提取公因子的方法，将 $g_1(x)$ 和 $g_2(x)$ 分别写成有理数乘以本原多项式的形式。于是有
$$g_1(x) = \frac{a_1}{b_1}f_1(x), \quad g_2(x) = \frac{a_2}{b_2}f_2(x),\quad f(x)=\frac{a_1a_2}{b_1b_2}f_1(x)f_2(x).$$

\item  因为本原多项式的乘积仍是本原多项式，所以 $f_1(x)f_2(x)$ 是本原多项式。

\item  因为 $f(x)$ 是整系数多项式，所以 $\frac{a_1a_2}{b_1b_2}$ 是一个整数。

\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.8.2.(Eisenstein判别法) 设 $f(x)=a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ 是一个整系数多项式。设存在一个素数 $p$ 使得
$$p\nmid a_n, \quad p\mid a_{n-1},\cdots, p\mid a_1,\quad p\mid a_0,\quad p^2\nmid a_0, $$  
那么这个多项式在有理系数范围内是不可约的。}

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\begin{itemize}

\item  {\color{red}例子：证明下述多项式在有理数域上是不可约的，
\begin{eqnarray*}
f(x) &=& x^n-2, \\
f(x) &=& x^4+4x^3+8x^2+4x+2.
\end{eqnarray*}
}

\item  解答：取 $p=2$, 由Eisenstein判别法得。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：证明不存在整数 $a,b,c$ 使得下述等式成立：
\begin{eqnarray*}
x^3+2x^2+2x+2 = (x+a)(x^2+bx+c).
\end{eqnarray*}
}

\item  证明： 将等式右边展开，比较系数，找到矛盾。


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：证明不存在整数 $a,b,c,d$ 使得下述等式成立：
\begin{eqnarray*}
x^4+3x^3+3x^2+3x+3 = (x^2+ax+b)(x^2+cx+d).
\end{eqnarray*}
}

\item  证明： 将等式右边展开，比较系数，找到矛盾。


\end{itemize}

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%每页详细内容

\begin{itemize}

\item  证明Eisenstein判别法：

\begin{enumerate}
\item  设整系数多项式 $f(x)$ 在有理系数范围内可约，则它可以写成两个次数较低的整系数多项式的乘积， 设 $f(x)=g(x)h(x)$. 设 
\begin{eqnarray*}
f(x) &=& a_0+a_1x+\cdots+a_nx^n, \\
g(x) &=& b_0+b_1x+\cdots+b_kx^k, \\
h(x) &=& c_0+c_1x+\cdots+c_mx^m. 
\end{eqnarray*}

\item  由 $a_0=b_0c_0$, $p\mid a_0$, $p^2\nmid a_0$ 可知 $p\mid b_0, p\nmid c_0$ 或 $p\nmid b_0, p\mid c_0$. 
\item  由 $a_n=b_kc_m$, $p\nmid a_n$ 想办法得出矛盾。
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：设 $p$ 是一个素数。判断下述多项式在有理数域上是否可约，
$$f(x)=x^{p-1}+x^{p-2}+\cdots+x+1.$$
}

\item  解答：
\begin{enumerate}
\item  无法直接使用 Eisenstein 判别法。
\item  取变量代换 $x=y+1$, 可得
$$f(x)=f(y+1)=y^{p-1}+\binom{p}{1}y^{p-2}+\cdots+\binom{p}{p-2}y+\binom{p}{p-1}=:g(y).$$
由 Eisenstein 判别法可知 $g(y)$ 是一个关于 $y$ 的不可约多项式。
\item  因此 $f(x)$ 也是一个关于 $x$ 的不可约多项式。
\end{enumerate}


\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：证明下述多项式不能写成两个次数更低的整系数多项式的乘积，
$$f(x)=x^6+x^5+x^4+x^3+x^2+x+1.$$
}

\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：在整系数多项式环中分解因式，
$$f(x)=x^5+x^4+x^3+x^2+x+1.$$
}

\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}定义：一个有理系数多项式 $f(x)$, 如果有一个有理数 $c$ 使得 $f(c)=0$, 则称 $c$ 为 $f(x)$ 的一个有理根。}

\item  {\color{red}问题：求下述多项式的有理根，
\begin{eqnarray*}
f(x) = (2x-3)(x^2-2) = 2x^3-3x^2-4x+6.
\end{eqnarray*}
}

\item  解答： $x=\frac{3}{2}$ 是唯一的有理根。

\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}定理2.8.3. 设有整系数多项式 $f(x)=a_0x^n+a_1x^{n-1}+\cdots+a_n$, 若有理数 $\frac{u}{v}$ 是 $f(x)$ 的一个根，其中 $u,v$ 为互素的整数，则
\begin{enumerate}
\item  {\color{red}$v\mid a_0, \quad u\mid a_n$.}
\item  {\color{red}$f(x)=(vx-u)g(x)$, 其中 $g(x)$ 是一个整系数多项式。}
\end{enumerate}
}

\item  证明： 
\begin{enumerate}
\item  因为 $\frac{u}{v}$ 是一个根，所以 $f(x)=(x-\frac{u}{v})q(x)$, 其中 $q(x)$ 是有理系数多项式。
\item  改写成 $f(x)=(vx-u)g(x)$, 因为 $vx-u$ 是本原多项式，而 $f(x)$ 是整系数多项式，所以 $g(x)$ 也是整系数多项式。
\item  设 $g(x)=b_0x^{n-1}+b_1x^{n-2}+\cdots+b_{n-1}$. 比较系数可知 $v\mid a_0$ 且 $u\mid a_n$.
\end{enumerate}

\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：求多项式 $f(x)=3x^4+5x^3+x^2+5x-2$ 的有理根。}

\item  解答：
\begin{enumerate}
\item  根据上述定理，可能的有理根一定形如 $\frac{u}{v}$, 其中 $v\mid 3, u\mid 2$. 
\item  于是所有可能为 $\pm 1,\quad \pm 2,\quad  \pm \frac{1}{3},\quad  \pm\frac{2}{3}.$
\item  逐个代入测试，可得有理根为 $-2$ 和 $\frac{1}{3}$. 
\item  最后用长除法或综合除法判断每个有理根的重数，都是单根。
\end{enumerate}

\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：设 $n$ 次多项式 $f(x)=a_0x^n + a_1x^{n-1}+\cdots+a_{n-1}x+a_0$ 的根是 $x_1,x_2,\cdots,x_n$. }
\begin{enumerate}
\item  {\color{red}求以 $cx_1,cx_2,\cdots,cx_n$ 为根的多项式。}
\item  {\color{red}求以 $1/x_1,1/x_2,\cdots,1/x_n$ 为根的多项式。}
\end{enumerate}


\item  解答：考虑韦达定理。


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题： 给出实系数的四次多项式在实数域上的所有不同类型的典型分解式。}

\item  解答：实系数的不可约多项式是一次多项式或者二次多项式。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题： 在复数和实数域上，将多项式 $x^n-2$ 分解因式。}

\item  解答：先考虑将 $x^2-1, x^3-1, x^4-1$ 分解因式。



\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题： 证明数域 $F$ 上的任意一个不可约多项式在复数域内没有重根。}

\item  解答：考虑导数多项式。


\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：证明下述多项式在有理数域上是不可约的，}
\begin{enumerate}
\item  {\color{red}$f(x)=x^4-2x^3+8x-10$. }
\item  {\color{red}$f(x)=2x^5+18x^4+6x^2+6$.} 
\item  {\color{red}$f(x)=x^4-2x^3+2x-3$. }
\item  {\color{red}$f(x)=x^6+x^3+1$. }
\end{enumerate}

\item  解答：使用 Eisenstein 判别法。必要时试试变量代换。


\end{itemize}

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\begin{itemize}


\item  {\color{red}问题：设 $p_1,p_2,p_3$ 是互不相同的素数，设整数 $n\ge 2$. 使用 Eisenstein 判别法证明 $\sqrt[n]{p_1p_2p_3}$ 是无理数。 }

\item  解答：证明多项式 $f(x) = x^n - p_1p_2p_3$ 是 $\mathbb{Q}[x]$ 中的不可约多项式。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：设 $f(x)$ 是一个整系数多项式。证明：如果 $f(0)$ 和 $f(1)$ 都是奇数，那么 $f(x)$ 没有整数根。}

\item  解答：假设 $f(x)$ 有整数根。考虑余数定理。

\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：求下述多项式的有理根，}
\begin{enumerate}
\item  {\color{red}$f(x)=x^3-6x^2+15x-14$. }
\item  {\color{red}$f(x)=4x^4-7x^2-5x-1$. }
\item  {\color{red}$f(x)=x^5-x^4-\frac{5}{2}x^3+2x^2-\frac{1}{2}x-3$. }
\end{enumerate}

\item  解答：整系数多项式的有理根（分子分母互素）有个特征，分母是最高次项系数的因子，分子是常数项的因子。这样就可以用枚举法来代入验证。
\begin{enumerate}
\item  $x=2$. 
\item  $x=-1/2$. 
\item  先化成整系数多项式。$x=-1, x=2$. 
\end{enumerate}

\end{itemize}

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%\item  
%\item  
%\item  
%\item  
%\item  证明 $x^5+ax+b=0$ 的求根公式。 
%\item  证明两个本原多项式的乘积仍然是一个本原多项式。
%\item  证明多项式 $f(x)=x^{2022}-2022$ 是不可约的整系数多项式，即不能写成两个次数较低的整系数多项式的乘积。
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There have also been some really effective metaphors used to encapsulate the experience of doing mathematics. Famously Andrew Wiles said:

“Perhaps I could best describe my experience of doing mathematics in terms of entering a dark mansion. You go into the first room and it’s dark, completely dark. You stumble around, bumping into the furniture. Gradually, you learn where each piece of furniture is. And finally, after six months or so, you find the light switch and turn it on. Suddenly, it’s all illuminated and you can see exactly where you were. Then you enter the next dark room…”

But perhaps my favourite metaphor for the process of being a mathematician was encapsulated in this quote from 2014 Fields Medalist Maryam Mirzakhani:

“There are times when I feel like I’m in a big forest and don’t know where I’m going. But then somehow I come to the top of a hill and can see everything more clearly. When that happens, it’s really exciting.”


